- #1

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**How do i read a frequency graph??**

If i have. a time required of 5-9 and a a frequency of 5. how do i find the average? do i do 9-4*6 or 9*5+5*9/5 ?? thanks.

- Thread starter Sirsh
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- #1

- 267

- 10

If i have. a time required of 5-9 and a a frequency of 5. how do i find the average? do i do 9-4*6 or 9*5+5*9/5 ?? thanks.

- #2

Mark44

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I don't think you have given a clear explanation of what you are given, but I'll do my best to explain what I think you are asking. Also, I don't think that either of the values you gave is correct. 9 - 4*6 = 9 - 24 = -15, which doesn't make any sense. If you meant (9 - 4)*6, that's 5*6 = 30, and I don't see that this number has anything to do with anything. For the other expression, 9*5+5*9/5 = 45 + 9 = 54, but that doesn't have anything to do with anything, either.If i have. a time required of 5-9 and a a frequency of 5. how do i find the average? do i do 9-4*6 or 9*5+5*9/5 ?? thanks.

I believe that the time required, 5 - 9, means a time interval of 5 to 9 time units of some kind, and the frequency of 5 means that 5 times were between 5 and 9 (sec? min? hours?).

The average time would be (5 + 9)/2 = 7

- #3

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Sorry i meant.

|Time required to complete test| Frequency | Midpoint |

| 5-9 | 6 | 7 |

| 10−14 | 12 | 12 |

| 15−19 | 25 | 17 |

| 20−24 | 34 | 22 |

| 25−29 | 16 | 27 |

| 30−34 | 7 | 32 |

What would be the mean of this? sorry.

Would i just do total of (frequency * midpoint)/amount scores?

- #4

Mark44

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Something like that. For each category, multiply the frequency by the midpoint of the category. Add all these products and divide by the total frequencies. IOW,

(f

- #5

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Thank you. Do you know how i'd go around getting the stadard deviation? Is it root(x-mean)^2/n?

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